Closed form for a conditional trigonometric series
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Do you know an easy way to prove the following $forall~L$?
$sumlimits_scriptstyle k = 1,~k ne Latop
scriptstyle ~l = 1,~l ne L^N cos left( frac2pi Nn(k -
l) right) - 2sumlimits_k = 1,~k ne L^N cos left( frac2pi Nn(k
- L) right) + 1 = left{ beginarrayl
(N - 2)^2rm~~if~n = 0\
4rm~~~if~n ne 0
endarray right.$
All parameters are integer numbers.
sequences-and-series trigonometry closed-form trigonometric-series
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up vote
0
down vote
favorite
Do you know an easy way to prove the following $forall~L$?
$sumlimits_scriptstyle k = 1,~k ne Latop
scriptstyle ~l = 1,~l ne L^N cos left( frac2pi Nn(k -
l) right) - 2sumlimits_k = 1,~k ne L^N cos left( frac2pi Nn(k
- L) right) + 1 = left{ beginarrayl
(N - 2)^2rm~~if~n = 0\
4rm~~~if~n ne 0
endarray right.$
All parameters are integer numbers.
sequences-and-series trigonometry closed-form trigonometric-series
Incorporate back the terms associated to $k=L$ and $l=L$, then exploit $costheta=textRe(e^itheta)$ and the formula for the sum of geometric progressions.
â Jack D'Aurizioâ¦
Aug 10 at 20:52
Thank you for your comments. I like Jack D'Aurizio 's approach. This question comes from here: https://math.stackexchange.com/questions/2878067/proof-that-certain-sums-of-complex-numbers-in-a-unitary-matrix-produce-the-sam
â HS TQ
Aug 10 at 21:02
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Do you know an easy way to prove the following $forall~L$?
$sumlimits_scriptstyle k = 1,~k ne Latop
scriptstyle ~l = 1,~l ne L^N cos left( frac2pi Nn(k -
l) right) - 2sumlimits_k = 1,~k ne L^N cos left( frac2pi Nn(k
- L) right) + 1 = left{ beginarrayl
(N - 2)^2rm~~if~n = 0\
4rm~~~if~n ne 0
endarray right.$
All parameters are integer numbers.
sequences-and-series trigonometry closed-form trigonometric-series
Do you know an easy way to prove the following $forall~L$?
$sumlimits_scriptstyle k = 1,~k ne Latop
scriptstyle ~l = 1,~l ne L^N cos left( frac2pi Nn(k -
l) right) - 2sumlimits_k = 1,~k ne L^N cos left( frac2pi Nn(k
- L) right) + 1 = left{ beginarrayl
(N - 2)^2rm~~if~n = 0\
4rm~~~if~n ne 0
endarray right.$
All parameters are integer numbers.
sequences-and-series trigonometry closed-form trigonometric-series
asked Aug 10 at 20:40
HS TQ
375
375
Incorporate back the terms associated to $k=L$ and $l=L$, then exploit $costheta=textRe(e^itheta)$ and the formula for the sum of geometric progressions.
â Jack D'Aurizioâ¦
Aug 10 at 20:52
Thank you for your comments. I like Jack D'Aurizio 's approach. This question comes from here: https://math.stackexchange.com/questions/2878067/proof-that-certain-sums-of-complex-numbers-in-a-unitary-matrix-produce-the-sam
â HS TQ
Aug 10 at 21:02
add a comment |Â
Incorporate back the terms associated to $k=L$ and $l=L$, then exploit $costheta=textRe(e^itheta)$ and the formula for the sum of geometric progressions.
â Jack D'Aurizioâ¦
Aug 10 at 20:52
Thank you for your comments. I like Jack D'Aurizio 's approach. This question comes from here: https://math.stackexchange.com/questions/2878067/proof-that-certain-sums-of-complex-numbers-in-a-unitary-matrix-produce-the-sam
â HS TQ
Aug 10 at 21:02
Incorporate back the terms associated to $k=L$ and $l=L$, then exploit $costheta=textRe(e^itheta)$ and the formula for the sum of geometric progressions.
â Jack D'Aurizioâ¦
Aug 10 at 20:52
Incorporate back the terms associated to $k=L$ and $l=L$, then exploit $costheta=textRe(e^itheta)$ and the formula for the sum of geometric progressions.
â Jack D'Aurizioâ¦
Aug 10 at 20:52
Thank you for your comments. I like Jack D'Aurizio 's approach. This question comes from here: https://math.stackexchange.com/questions/2878067/proof-that-certain-sums-of-complex-numbers-in-a-unitary-matrix-produce-the-sam
â HS TQ
Aug 10 at 21:02
Thank you for your comments. I like Jack D'Aurizio 's approach. This question comes from here: https://math.stackexchange.com/questions/2878067/proof-that-certain-sums-of-complex-numbers-in-a-unitary-matrix-produce-the-sam
â HS TQ
Aug 10 at 21:02
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
for $n=0$ we have
$$ (N-1)^2-2(N-1)+1 =(N-2)^2 $$
If $nne0$ , using the symmetry $ cos(pi - x) = -cos(x)$ (or any other method) you should be able to convince yourself that
$$ sumlimits_k = 1^N cos left( frac2pi Nn(k
- L) right) =0
$$
For all integers $N, L, n ne 0$.
So
$$sumlimits_k = 1,~k ne L^N cos left( frac2pi Nn(k
- L) right) = 0 - cos left( frac2pi Nn(L
- L) right) =-1
$$
And
$$sumlimits_scriptstyle k = 1,~k ne Latop
scriptstyle ~l = 1,~l ne L^N cos left( frac2pi Nn(k -
l) right)
\ = sumlimits_scriptstyle k = 1,~k ne L^N - cos left( frac2pi Nn(k -
L) right)
\ = cos left( frac2pi Nn(L
- L) right)
=1$$
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
for $n=0$ we have
$$ (N-1)^2-2(N-1)+1 =(N-2)^2 $$
If $nne0$ , using the symmetry $ cos(pi - x) = -cos(x)$ (or any other method) you should be able to convince yourself that
$$ sumlimits_k = 1^N cos left( frac2pi Nn(k
- L) right) =0
$$
For all integers $N, L, n ne 0$.
So
$$sumlimits_k = 1,~k ne L^N cos left( frac2pi Nn(k
- L) right) = 0 - cos left( frac2pi Nn(L
- L) right) =-1
$$
And
$$sumlimits_scriptstyle k = 1,~k ne Latop
scriptstyle ~l = 1,~l ne L^N cos left( frac2pi Nn(k -
l) right)
\ = sumlimits_scriptstyle k = 1,~k ne L^N - cos left( frac2pi Nn(k -
L) right)
\ = cos left( frac2pi Nn(L
- L) right)
=1$$
add a comment |Â
up vote
1
down vote
accepted
for $n=0$ we have
$$ (N-1)^2-2(N-1)+1 =(N-2)^2 $$
If $nne0$ , using the symmetry $ cos(pi - x) = -cos(x)$ (or any other method) you should be able to convince yourself that
$$ sumlimits_k = 1^N cos left( frac2pi Nn(k
- L) right) =0
$$
For all integers $N, L, n ne 0$.
So
$$sumlimits_k = 1,~k ne L^N cos left( frac2pi Nn(k
- L) right) = 0 - cos left( frac2pi Nn(L
- L) right) =-1
$$
And
$$sumlimits_scriptstyle k = 1,~k ne Latop
scriptstyle ~l = 1,~l ne L^N cos left( frac2pi Nn(k -
l) right)
\ = sumlimits_scriptstyle k = 1,~k ne L^N - cos left( frac2pi Nn(k -
L) right)
\ = cos left( frac2pi Nn(L
- L) right)
=1$$
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
for $n=0$ we have
$$ (N-1)^2-2(N-1)+1 =(N-2)^2 $$
If $nne0$ , using the symmetry $ cos(pi - x) = -cos(x)$ (or any other method) you should be able to convince yourself that
$$ sumlimits_k = 1^N cos left( frac2pi Nn(k
- L) right) =0
$$
For all integers $N, L, n ne 0$.
So
$$sumlimits_k = 1,~k ne L^N cos left( frac2pi Nn(k
- L) right) = 0 - cos left( frac2pi Nn(L
- L) right) =-1
$$
And
$$sumlimits_scriptstyle k = 1,~k ne Latop
scriptstyle ~l = 1,~l ne L^N cos left( frac2pi Nn(k -
l) right)
\ = sumlimits_scriptstyle k = 1,~k ne L^N - cos left( frac2pi Nn(k -
L) right)
\ = cos left( frac2pi Nn(L
- L) right)
=1$$
for $n=0$ we have
$$ (N-1)^2-2(N-1)+1 =(N-2)^2 $$
If $nne0$ , using the symmetry $ cos(pi - x) = -cos(x)$ (or any other method) you should be able to convince yourself that
$$ sumlimits_k = 1^N cos left( frac2pi Nn(k
- L) right) =0
$$
For all integers $N, L, n ne 0$.
So
$$sumlimits_k = 1,~k ne L^N cos left( frac2pi Nn(k
- L) right) = 0 - cos left( frac2pi Nn(L
- L) right) =-1
$$
And
$$sumlimits_scriptstyle k = 1,~k ne Latop
scriptstyle ~l = 1,~l ne L^N cos left( frac2pi Nn(k -
l) right)
\ = sumlimits_scriptstyle k = 1,~k ne L^N - cos left( frac2pi Nn(k -
L) right)
\ = cos left( frac2pi Nn(L
- L) right)
=1$$
answered Aug 10 at 21:30
WW1
6,4821712
6,4821712
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Incorporate back the terms associated to $k=L$ and $l=L$, then exploit $costheta=textRe(e^itheta)$ and the formula for the sum of geometric progressions.
â Jack D'Aurizioâ¦
Aug 10 at 20:52
Thank you for your comments. I like Jack D'Aurizio 's approach. This question comes from here: https://math.stackexchange.com/questions/2878067/proof-that-certain-sums-of-complex-numbers-in-a-unitary-matrix-produce-the-sam
â HS TQ
Aug 10 at 21:02